Problem: Simplify the following expression and state the conditions under which the simplification is valid. You can assume that $q \neq 0$. $n = \dfrac{q + 2}{q^2 + 3q + 2} \div \dfrac{q - 9}{7q + 7} $
Dividing by an expression is the same as multiplying by its inverse. $n = \dfrac{q + 2}{q^2 + 3q + 2} \times \dfrac{7q + 7}{q - 9} $ First factor the quadratic. $n = \dfrac{q + 2}{(q + 1)(q + 2)} \times \dfrac{7q + 7}{q - 9} $ Then factor out any other terms. $n = \dfrac{q + 2}{(q + 1)(q + 2)} \times \dfrac{7(q + 1)}{q - 9} $ Then multiply the two numerators and multiply the two denominators. $n = \dfrac{ (q + 2) \times 7(q + 1) } { (q + 1)(q + 2) \times (q - 9) } $ $n = \dfrac{ 7(q + 2)(q + 1)}{ (q + 1)(q + 2)(q - 9)} $ Notice that $(q + 2)$ and $(q + 1)$ appear in both the numerator and denominator so we can cancel them. $n = \dfrac{ 7(q + 2)\cancel{(q + 1)}}{ \cancel{(q + 1)}(q + 2)(q - 9)} $ We are dividing by $q + 1$ , so $q + 1 \neq 0$ Therefore, $q \neq -1$ $n = \dfrac{ 7\cancel{(q + 2)}\cancel{(q + 1)}}{ \cancel{(q + 1)}\cancel{(q + 2)}(q - 9)} $ We are dividing by $q + 2$ , so $q + 2 \neq 0$ Therefore, $q \neq -2$ $n = \dfrac{7}{q - 9} ; \space q \neq -1 ; \space q \neq -2 $